[EM] Double Defeat, Hare

[Chris Benham]

On April 9 last year I suggested the "Double Defeat, Hare" method 
(nominating it as an alternative in a poll).

> *Voters strictly rank from the top however many candidates they wish and also may specify an approval cutoff.
>
> Default approval is only goes to top-ranked candidates.
>
> All candidates that are pairwise beaten by a more approved candidate are disqualified.
>
> If that leaves more than one qualified (i.e. not disqualified) candidate, commence eliminations according to Hare rules until only one qualified candidate remains.*


I now withdraw that not-so-clever idea because of this scenario:

43 A|
03 A<B|
44 B>C|  (sincere is B| or B|>A or B>A|)
10 C|>A

A is both the normal Hare winner and the sincere Condorcet winner.

Approvals:   C 54     B 47    A 46.

C>A 54-47,     A>B 56-44,     B>C 47-10.

Only A is disqualified by Double Defeat  and  C is eliminated by the 
Hare rule leaving B the winner, rewarding the outrageous Burial strategists.

My favourite Condorcet method, Margins-Sorted Approval(explicit) would 
punish them by electing C   (as would the not-too-bad Smith//Approval).

It would first look at the BA pair because they are adjacent to each 
other in the approval order and have a smaller margin of difference in 
their approval scores (47-46=1) than the CB adjacent pair (54-47=7) and 
notice that they are pairwise (by the rankings) out of order
and so flip that order, making it  C>A>B.  Neither of the two adjacent 
pairs are now pairwise out of order so that order is final  and C is on 
top of it so C wins.

Smith//Approval(explicit)  sees that all three candidates are in the 
Smith set and so elects the one with the most approval.

Hare, being completely immune to Burial strategy (because it meets 
Later-no-Help) elects (in this example)  the sincere Condorcet Winner.

Chris  Benham
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